Solution — 2022 Question 3 Quadratic Equations & Inequalities

(a)

\begin{matrix} F E & = y - 2 (3 x) \\ = y - 6 x \end{matrix}

\begin{matrix} G A & = 12 x - 4 x \\ = 8 x \end{matrix}

By Pythagoras’ Theorem,

\begin{matrix} G F & = \sqrt{{(3 x)}^{2} + {(4 x)}^{2}} \\ = 5 x \end{matrix}

Considering the perimeter,

\begin{matrix} F E + 2 G F + 2 G A + A B & = 80 \\ y - 6 x + 2 (5 x) + 2 (8 x) + y & = 80 \\ 2 y + 20 x & = 80 \\ y + 10 x & = 40 ∎ & (1) \end{matrix}

(b)

Considering the area,

\begin{matrix} Area of rectangle A B C D - 2 (Area of triangle D F G) & = 432 \\ y (12 x) - 2 (\frac{1}{2}) (4 x) (3 x) & = 432 \\ 12 y x - 12 x^{2} & = 432 \\ y x - x^{2} & = 36 & (2) \end{matrix}

From (1),

\begin{matrix} y + 10 x & = 40 \\ y & = 40 - 10 x & (3) \end{matrix}

Substituting (3) into (2):

\begin{matrix} (40 - 10 x) x - x^{2} & = 36 \\ 40 x - 10 x^{2} - x^{2} & = 36 \\ 40 x - 11 x^{2} & = 36 \\ 11 x^{2} - 40 x + 36 & = 0 \\ (11 x - 18) (x - 2) & = 0 \end{matrix}

$x = \frac{18}{11} ∎ or x = 2 ∎ .$

When $x = \frac{18}{11} :$

\begin{matrix} y & = 40 - 10 (\frac{18}{11}) \\ = \frac{260}{11} ∎ \end{matrix}

When $x = 2 :$

\begin{matrix} y & = 40 - 10 (2) \\ = 20 ∎ \end{matrix}